Prove converging sequence

Question

Problem:

{$x_n$} is a sequence of positive real numbers that satisfies $x_n\leq\frac {n+1}{n^2}$ for all $n\in\Bbb N$. Prove that the sequence $\{x_n(-1)^n\}$ is convergent.

My thoughts:

I notice that $x_n$ goes to 0 via the squeeze lemma. And since $(-1)^n$ is being multiplied by 0, the whole sequence will just converge to zero. Am I correct in thinking this way? If not, then how would I go about proving this?

Answer 1

We shall use the following result.

Result. $\lim_{n\to\infty}|a_n|=0$ iff $\lim_{n\to\infty}a_n=0$.

In your question, we get $$|x_n(-1)^n|=|x_n|=x_n\quad\forall n\in\Bbb N.$$

Now, $$0

Answer 2

Let $ z_n:=(-1)^nx_n$. Then $|z_n|=|x_n|=x_n$. Hence $|z_n| \to 0$ and therefore $z_n \to 0$